Continuity Questions in English

(C) Since $f(x)$ is continuous at $x=\frac{\pi}{2}$,we have $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.
Given $f(\frac{\pi}{2}) = 2024$.
Now,$\lim_{x \to \frac{\pi}{2}} \frac{2k \cos x}{\pi-2x} = 2024$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
$\lim_{h \to 0} \frac{2k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{2k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-2k \sin h}{-2h} = k \lim_{h \to 0} \frac{\sin h}{h} = k(1) = k$.
Equating the limit to the function value,we get $k = 2024$.

(A) For a function $f(\alpha)$ to be continuous at $\alpha=0$,the limit of the function as $\alpha \to 0$ must equal the value of the function at $\alpha=0$.
Given $f(0) = k$.
We need to evaluate $\lim_{\alpha \to 0} f(\alpha) = \lim_{\alpha \to 0} \frac{1-\cos 6 \alpha}{36 \alpha^2}$.
Using the trigonometric identity $1-\cos \theta = 2 \sin^2(\theta/2)$,we have $1-\cos 6 \alpha = 2 \sin^2(3 \alpha)$.
Substituting this into the limit: $\lim_{\alpha \to 0} \frac{2 \sin^2(3 \alpha)}{36 \alpha^2} = \lim_{\alpha \to 0} \frac{2}{36} \left( \frac{\sin 3 \alpha}{\alpha} \right)^2$.
Using the standard limit $\lim_{x \to 0} \frac{\sin ax}{x} = a$,we get $\lim_{\alpha \to 0} \frac{1}{18} (3)^2 = \frac{9}{18} = \frac{1}{2}$.
Since the function is continuous,$k = 1/2$.

(D) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = \frac{\pi}{2}$ must be equal.
$1$. Value of the function at $x = \frac{\pi}{2}$:
$f(\frac{\pi}{2}) = k(\frac{\pi}{2}) + 1$
$2$. Left-hand limit $(LHL)$:
$\lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} (kx + 1) = k(\frac{\pi}{2}) + 1$
$3$. Right-hand limit $(RHL)$:
$\lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \sin x = \sin(\frac{\pi}{2}) = 1$
For continuity,$LHL = RHL$:
$k(\frac{\pi}{2}) + 1 = 1$
$k(\frac{\pi}{2}) = 0$
$k = 0$
Thus,the correct option is $D$.

(B) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the limit of $f(x)$ as $x \to \frac{\pi}{2}$ must equal $f(\frac{\pi}{2})$.
Given $f(\frac{\pi}{2}) = 3$.
We calculate the limit: $\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
Substituting this into the limit:
$\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$.
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$,the limit is $\frac{k}{2}$.
Equating the limit to the function value: $\frac{k}{2} = 3 \implies k = 6$.

(C) For a function $f(x)$ to be continuous at $x = a$,the limit $\lim_{x \to a} f(x)$ must equal $f(a)$.
Given $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = \frac{1}{2}$.
Now,calculate the limit: $\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
Substituting this into the limit: $\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$.
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$,the limit is $\frac{k}{2}$.
Equating this to $f(\frac{\pi}{2}) = \frac{1}{2}$,we get $\frac{k}{2} = \frac{1}{2}$,which implies $k = 1$.

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit,right-hand limit,and the value of the function at $x = a$ must be equal.
Specifically,$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$.
Left-hand limit at $x = 3$: $\lim_{x \to 3^-} (cx + 1) = 3c + 1$.
Right-hand limit at $x = 3$: $\lim_{x \to 3^+} (dx + 3) = 3d + 3$.
Since $f$ is continuous at $x = 3$,we have $3c + 1 = 3d + 3$.
Rearranging the terms: $3c - 3d = 3 - 1$.
$3(c - d) = 2$.
$c - d = \frac{2}{3}$.
We need to find $d - c$,so $d - c = -(c - d) = -\frac{2}{3}$.
Therefore,the correct option is $C$.

(A) For a function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \to 0$ must be equal to $f(0)$.
Given $f(0) = 1$.
We calculate $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 5x \tan kx}{x^2}$.
Using the standard limits $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,we rewrite the expression:
$\lim_{x \to 0} \left( \frac{\sin 5x}{5x} \times 5 \right) \times \left( \frac{\tan kx}{kx} \times k \right) = 1 \times 5 \times 1 \times k = 5k$.
Since the function is continuous,$5k = 1$.
Therefore,$k = 1/5$.

(C) For a function $f(x)$ to be continuous at $x = 0$,the limit of the function as $x \to 0$ must be equal to the value of the function at $x = 0$.
That is,$\lim_{x \to 0} f(x) = f(0)$.
Given $f(0) = k$.
Now,calculate the limit: $\lim_{x \to 0} \frac{\tan 4x \times \cos 3x}{x}$.
We can rewrite this as: $\lim_{x \to 0} \left( \frac{\tan 4x}{x} \times \cos 3x \right)$.
Multiply and divide by $4$ to use the standard limit $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$:
$\lim_{x \to 0} \left( 4 \times \frac{\tan 4x}{4x} \times \cos 3x \right)$.
Applying the limits: $4 \times 1 \times \cos(3 \times 0) = 4 \times 1 \times \cos(0) = 4 \times 1 \times 1 = 4$.
Therefore,$k = 4$.

(D) Given the function $f(x) = [x]$.
We know that the greatest integer function $[x]$ is discontinuous at all integer values of $x$.
It is continuous at all non-integer values.
Among the given options,$4$,$-2$,and $11$ are integers,while $1.5$ is a non-integer.
Therefore,$f(x)$ is continuous at $x = 1.5$.

(A) For a function to be continuous at $x = 0$,the limit of the function as $x \to 0$ must be equal to the value of the function at $x = 0$.
Given $f(x) = \frac{3 \sin(\pi x)}{5x}$ for $x \neq 0$ and $f(0) = 2K$.
We know that $\lim_{x \to 0} \frac{\sin(\theta x)}{x} = \theta$.
Therefore,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3 \sin(\pi x)}{5x} = \frac{3}{5} \lim_{x \to 0} \frac{\sin(\pi x)}{x} = \frac{3}{5} \times \pi = \frac{3\pi}{5}$.
Since the function is continuous at $x = 0$,we have $\lim_{x \to 0} f(x) = f(0)$.
So,$\frac{3\pi}{5} = 2K$.
Solving for $K$,we get $K = \frac{3\pi}{10}$.

(B) For the function to be continuous at $x=0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the standard limit $\lim_{x \to 0} \frac{\log(1+kx)}{x} = k$,we have:
$\lim_{x \to 0} \frac{\log(1+ax) - \log(1-bx)}{x} = \lim_{x \to 0} \left( \frac{\log(1+ax)}{x} - \frac{\log(1-bx)}{x} \right)$
$= \lim_{x \to 0} \frac{\log(1+ax)}{x} - \lim_{x \to 0} \frac{\log(1-bx)}{x}$
$= a - (-b) = a + b$.
Thus,the value to be assigned is $a+b$.

(C) Given that the function $f(x)$ is continuous at $x = 2$,the left-hand limit ($L$.$H$.$L$.) must equal the right-hand limit ($R$.$H$.$L$.).
$L$.$H$.$L$. = $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} Kx^2 = K(2)^2 = 4K$.
$R$.$H$.$L$. = $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} 3 = 3$.
Since the function is continuous,$4K = 3$.
Therefore,$K = \frac{3}{4}$.

(B) To check the continuity at $x=0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
For $LHL$: $\lim_{x \to 0^-} f(x) = \lim_{h \to 0} \frac{e^{1/(0-h)}-1}{e^{1/(0-h)}+1} = \lim_{h \to 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$. Since $e^{-1/h} \to 0$ as $h \to 0^+$,$LHL$ $= \frac{0-1}{0+1} = -1$.
For $RHL$: $\lim_{x \to 0^+} f(x) = \lim_{h \to 0} \frac{e^{1/h}-1}{e^{1/h}+1} = \lim_{h \to 0} \frac{1-e^{-1/h}}{1+e^{-1/h}} = \frac{1-0}{1+0} = 1$.
Since $LHL$ $\neq$ $RHL$,the limit does not exist at $x=0$.
Therefore,the function is not continuous at $x=0$.

(D) Given the function $f(x) = \cot x$.
We can write this as $f(x) = \frac{\cos x}{\sin x}$.
$A$ rational function is discontinuous where its denominator is equal to zero.
Therefore,$f(x)$ is discontinuous where $\sin x = 0$.
The general solution for $\sin x = 0$ is $x = n\pi$,where $n \in Z$.
Thus,the function $f(x) = \cot x$ is discontinuous on the set $\{x = n\pi ; n \in Z\}$.

(D) For the function $f(x)$ to be continuous at $x=0$,the limit of $f(x)$ as $x \to 0$ must equal $f(0)$.
$\lim_{x \to 0} f(x) = f(0)$
$\lim_{x \to 0} \frac{\sin 3x}{e^{2x}-1} = k-2$
Divide the numerator and denominator by $x$:
$\lim_{x \to 0} \frac{\frac{\sin 3x}{x}}{\frac{e^{2x}-1}{x}} = k-2$
Using the standard limits $\lim_{x \to 0} \frac{\sin ax}{x} = a$ and $\lim_{x \to 0} \frac{e^{ax}-1}{x} = a$:
$\frac{3}{2} = k-2$
$k = \frac{3}{2} + 2$
$k = \frac{3+4}{2} = \frac{7}{2}$

(B) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ at $x=0$.
First,calculate the $LHL$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
Multiply the numerator and denominator by the conjugate $(\sqrt{1+kx} + \sqrt{1-kx})$:
$= \lim_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1+1} = k$.
Next,calculate the $RHL$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2x+1}{x-1} = \frac{2(0)+1}{0-1} = -1$.
Since the function is continuous at $x=0$,we set $LHL$ = $RHL$:
$k = -1$.

(A) For a function $f(x)$ to be continuous at $x=a$,the limit $\lim_{x \to a} f(x)$ must exist and be equal to $f(a)$.
Given $f(x) = \frac{\log_{e} x}{x-1}$ for $x \neq 1$ and $f(1) = k$.
We need to evaluate $\lim_{x \to 1} \frac{\log_{e} x}{x-1}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'$H$ôpital's Rule:
$\lim_{x \to 1} \frac{\frac{d}{dx}(\log_{e} x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = \frac{1}{1} = 1$.
Since the function is continuous at $x=1$,we must have $k = \lim_{x \to 1} f(x) = 1$.

(A) The greatest integer function $f(x) = [x]$ is defined as the greatest integer less than or equal to $x$.
It is a well-known property that the greatest integer function is discontinuous at every integer value $n \in \mathbb{Z}$.
Conversely,the function is continuous at all non-integer values.
Comparing the given options:
$A) 1.5$ (Non-integer)
$B) 4$ (Integer)
$C) 1$ (Integer)
$D) -2$ (Integer)
Since $1.5$ is not an integer,the function $f(x) = [x]$ is continuous at $x = 1.5$.

(C) Given,$f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$.
For any $a \neq 0$,consider a sequence of rational numbers $r_n \to a$ and a sequence of irrational numbers $i_n \to a$.
Then $\lim_{n \to \infty} f(r_n) = \lim_{n \to \infty} 0 = 0$ and $\lim_{n \to \infty} f(i_n) = \lim_{n \to \infty} i_n = a$.
Since $0 \neq a$ for $a \neq 0$,the limit $\lim_{x \to a} f(x)$ does not exist for $a \neq 0$.
At $x=0$,$\lim_{x \to 0} f(x) = 0$ because for any sequence $x_n \to 0$,$f(x_n)$ is either $x_n$ (if $x_n$ is irrational) or $0$ (if $x_n$ is rational),both of which approach $0$.
Since $f(0) = 0$,the function is continuous only at $x=0$.

(B) For a function $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x \to 2$ must equal $f(2)$.
Given $f(2) = 2$.
We evaluate the limit: $\lim_{x \to 2} \frac{x^2 - (a+2)x + a}{x-2}$.
Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ at $x = 2$ for the limit to be finite.
Substituting $x = 2$ in the numerator: $2^2 - (a+2)(2) + a = 0$.
$4 - 2a - 4 + a = 0$.
$-a = 0 \implies a = 0$.
Alternatively,using $L$'Hospital's rule: $\lim_{x \to 2} \frac{2x - (a+2)}{1} = 2(2) - a - 2 = 2 - a$.
Equating to $f(2)$: $2 - a = 2 \implies a = 0$.

(C) Given that the function $f(x)$ is continuous at $x=1$,the condition for continuity is $f(1) = \lim_{x \to 1} f(x)$.
From the definition of the function,$f(1) = k$.
Therefore,we need to evaluate the limit: $k = \lim_{x \to 1} \frac{\log x}{x-1}$.
Substituting $x=1$,we get the indeterminate form $\frac{0}{0}$.
Applying $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:
$k = \lim_{x \to 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1}$.
Evaluating the limit as $x \to 1$:
$k = \frac{1/1}{1} = 1$.
Thus,the value of $k$ is $1$.

(A) The function $f(x) = [x]$ is known as the Greatest Integer Function.
For any integer $n$,the left-hand limit is $\lim_{x \to n^-} f(x) = n-1$ and the right-hand limit is $\lim_{x \to n^+} f(x) = n$.
Since the left-hand limit is not equal to the right-hand limit at any integer $n$,the function is discontinuous at all integral values of $x$.
However,for any non-integral value $c$ (where $c$ is not an integer),the function is locally constant in a small neighborhood around $c$,making it continuous at all such points.
Therefore,the function $f(x) = [x]$ is continuous for all non-integral values of $x$.

(D) Given $f(x) = |x-2| + x$.
First,we check for continuity at $x=0$ and $x=2$.
At $x=0$: $f(0) = |0-2| + 0 = 2$. $\lim_{x \to 0^-} f(x) = |0-2| + 0 = 2$ and $\lim_{x \to 0^+} f(x) = |0-2| + 0 = 2$. Since $\lim_{x \to 0} f(x) = f(0)$,the function is continuous at $x=0$.
At $x=2$: $f(2) = |2-2| + 2 = 2$. $\lim_{x \to 2^-} f(x) = |2-2| + 2 = 2$ and $\lim_{x \to 2^+} f(x) = |2-2| + 2 = 2$. Since $\lim_{x \to 2} f(x) = f(2)$,the function is continuous at $x=2$.
Now,we check for differentiability. We can write $f(x)$ as:
$f(x) = \begin{cases} -(x-2) + x, & x < 2 \\ (x-2) + x, & x \geq 2 \end{cases} = \begin{cases} 2, & x < 2 \\ 2x-2, & x \geq 2 \end{cases}$.
For $x < 2$,$f'(x) = 0$. For $x > 2$,$f'(x) = 2$.
At $x=2$,$LHD$ $= \lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0^-} \frac{2-2}{h} = 0$.
$RHD$ $= \lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0^+} \frac{2(2+h)-2-2}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Since $LHD$ $\neq$ $RHD$,$f(x)$ is not differentiable at $x=2$.
At $x=0$,$f(x) = 2$ in the neighborhood,so $f'(0) = 0$. Thus,$f(x)$ is differentiable at $x=0$.
Therefore,the function is continuous at both $x=0$ and $x=2$.

(C) Given the function $f(x) = \begin{cases} \frac{1-\cos x}{x^{2}}, & x \neq 0 \\ k, & x=0 \end{cases}$.
Since the function is continuous at $x=0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0)$.
Therefore,$\lim_{x \rightarrow 0} \frac{1-\cos x}{x^{2}} = k$.
Using the limit formula $\lim_{x \rightarrow 0} \frac{1-\cos x}{x^{2}} = \frac{1}{2}$,we get:
$\lim_{x \rightarrow 0} \frac{2\sin^{2}(x/2)}{x^{2}} = \lim_{x \rightarrow 0} \frac{2\sin^{2}(x/2)}{4(x/2)^{2}} = \frac{2}{4} \cdot (1)^{2} = \frac{1}{2}$.
Thus,$k = \frac{1}{2}$.

(D) Given that $f(x)$ is continuous at $x=0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,$\lim_{x \to 0} \frac{1-\cos Kx}{x \sin x} = \frac{1}{2}$.
Using the trigonometric identity $1-\cos \theta = 2 \sin^2(\frac{\theta}{2})$,we get:
$\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x \sin x} = \frac{1}{2}$.
Dividing numerator and denominator by $x^2$,we get:
$\lim_{x \to 0} \frac{2 \left(\frac{\sin(Kx/2)}{x}\right)^2}{\frac{\sin x}{x}} = \frac{1}{2}$.
Multiplying and dividing by $(\frac{K}{2})^2$ in the numerator:
$\lim_{x \to 0} \frac{2 \cdot (\frac{K}{2})^2 \cdot (\frac{\sin(Kx/2)}{Kx/2})^2}{\frac{\sin x}{x}} = \frac{1}{2}$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we have:
$2 \cdot \frac{K^2}{4} \cdot \frac{1^2}{1} = \frac{1}{2}$.
$\frac{K^2}{2} = \frac{1}{2} \implies K^2 = 1$.
Thus,$K = \pm 1$.

(D) For a function $f(x)$ to be continuous at $x = 5$,the left-hand limit ($L$.$H$.$L$.) must equal the right-hand limit ($R$.$H$.$L$.) and the value of the function at $x = 5$.
First,calculate the value of the function at $x = 5$: $f(5) = 3(5) - 8 = 15 - 8 = 7$.
Next,calculate the $L$.$H$.$L$. as $x \to 5^-$: $\lim_{x \to 5^-} (3x - 8) = 3(5) - 8 = 7$.
Then,calculate the $R$.$H$.$L$. as $x \to 5^+$: $\lim_{x \to 5^+} (2k) = 2k$.
Since the function is continuous,we set $L$.$H$.$L$. = $R$.$H$.$L$.: $7 = 2k$.
Solving for $k$,we get $k = 7/2$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 5$.
First,consider the left-hand limit: $\lim_{x \to 0^-} \frac{(1+ax^2+bx^3)^{1/3} - (1-ax^2-bx^3)^{1/3}}{x^2}$.
Using the binomial expansion $(1+u)^{1/3} \approx 1 + \frac{1}{3}u - \frac{1}{9}u^2$,we get:
$\lim_{x \to 0^-} \frac{(1 + \frac{1}{3}(ax^2+bx^3)) - (1 + \frac{1}{3}(-ax^2-bx^3))}{x^2} = \lim_{x \to 0^-} \frac{\frac{2}{3}ax^2 + \frac{2}{3}bx^3}{x^2} = \frac{2}{3}a$.
Since the limit is $5$,we have $\frac{2}{3}a = 5 \implies a = \frac{15}{2}$.
Next,consider the right-hand limit: $\lim_{x \to 0^+} \frac{\tan 3x - \sin 3x}{bx^3}$.
Using Taylor series $\tan 3x \approx 3x + \frac{(3x)^3}{3}$ and $\sin 3x \approx 3x - \frac{(3x)^3}{6}$:
$\lim_{x \to 0^+} \frac{(3x + 9x^3) - (3x - \frac{27}{6}x^3)}{bx^3} = \lim_{x \to 0^+} \frac{9x^3 + 4.5x^3}{bx^3} = \frac{13.5}{b} = \frac{27}{2b}$.
Since the limit is $5$,we have $\frac{27}{2b} = 5 \implies b = \frac{27}{10}$.
Wait,re-evaluating the expansion: $\tan 3x - \sin 3x = \sin 3x (\sec 3x - 1) \approx (3x)(1 + \frac{(3x)^2}{2} - 1) = \frac{27x^3}{2}$.
So $\frac{27/2}{b} = 5 \implies b = \frac{27}{10}$.
Geometric mean of $a$ and $b$ is $\sqrt{ab} = \sqrt{\frac{15}{2} \times \frac{27}{10}} = \sqrt{\frac{3 \times 27}{2 \times 2}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.

(C) For option $C$,$f(x) = \begin{cases} \frac{x-1}{|x-1|+2(x-1)^2}, & x \neq 1 \\ 1, & x=1 \end{cases}$
For $x > 1$,$|x-1| = x-1$,so $f(x) = \frac{x-1}{(x-1)+2(x-1)^2} = \frac{1}{1+2(x-1)} = \frac{1}{2x-1}$.
For $x < 1$,$|x-1| = -(x-1)$,so $f(x) = \frac{x-1}{-(x-1)+2(x-1)^2} = \frac{1}{-1+2(x-1)} = \frac{1}{2x-3}$.
Now,$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{2x-1} = \frac{1}{2(1)-1} = 1$.
And $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{1}{2x-3} = \frac{1}{2(1)-3} = -1$.
Since $\lim_{x \to 1^+} f(x) \neq \lim_{x \to 1^-} f(x)$,the limit does not exist at $x=1$.
Therefore,$f(x)$ is discontinuous at $x=1$.

(A) Given $f(x)$ is continuous on $R$,it must be continuous at $x=0$ and $x=3$.
For continuity at $x=0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
$\lim_{x \to 0^-} \frac{x-|x|}{x} = \lim_{x \to 0^-} \frac{x-(-x)}{x} = \lim_{x \to 0^-} \frac{2x}{x} = 2$.
$\lim_{x \to 0^+} b\left(\frac{5x^2+a}{x^2-3x+2}\right) = b\left(\frac{a}{2}\right)$.
Equating them: $b\left(\frac{a}{2}\right) = 2 \implies ab = 4$.
For continuity at $x=3$: $\lim_{x \to 3^-} f(x) = f(3) = -14$.
Assuming the middle piece is valid up to $x=3$: $\lim_{x \to 3^-} b\left(\frac{5x^2+a}{x^2-3x+2}\right) = b\left(\frac{5(9)+a}{9-9+2}\right) = b\left(\frac{45+a}{2}\right) = -14$.
$b(45+a) = -28$.
Substituting $a = 4/b$: $b(45 + 4/b) = -28 \implies 45b + 4 = -28 \implies 45b = -32$.
Given the options,if we test $(a, b) = (2, -7/2)$,then $ab = -7 \neq 4$. There is a typo in the provided function definition. However,based on standard problem patterns,$(A)$ is the intended choice.

(C) Given the function $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$.
For the limit to exist at $x = 1$,we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$.
Calculating the left-hand limit: $\lim_{x \to 1^-} (1 + 2x^k) = 1 + 2(1)^k = 1 + 2 = 3$.
Calculating the right-hand limit: $\lim_{x \to 1^+} (kx) = k(1) = k$.
Equating the two limits: $3 = k$.
Since $k = 3$,the value of $k^2$ is $3^2 = 9$.

(D) For $f(x)$ to be right continuous at $x = -1$,we must have $\lim_{x \rightarrow -1^+} f(x) = f(-1)$.
Applying $L'H\hat{o}pital's$ rule to the limit $\lim_{x \rightarrow -1^+} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}$:
$\lim_{x \rightarrow -1^+} \frac{-\frac{1}{2\sqrt{\cos^{-1} x}} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)}{\frac{1}{2\sqrt{x+1}}} = \lim_{x \rightarrow -1^+} \frac{\sqrt{x+1}}{\sqrt{\cos^{-1} x} \sqrt{1-x} \sqrt{1+x}}$
$= \lim_{x \rightarrow -1^+} \frac{1}{\sqrt{\cos^{-1} x} \sqrt{1-x}} = \frac{1}{\sqrt{\cos^{-1}(-1)} \sqrt{1-(-1)}} = \frac{1}{\sqrt{\pi} \sqrt{2}} = \frac{1}{\sqrt{2\pi}}$.
Given $f(-1) = \frac{1}{\sqrt{\lambda \pi}}$,we equate: $\frac{1}{\sqrt{2\pi}} = \frac{1}{\sqrt{\lambda \pi}}$.
Squaring both sides gives $2\pi = \lambda \pi$,so $\lambda = 2$.

(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
Evaluating the limit: $\lim_{x \to 0} \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$.
This is a $1^{\infty}$ indeterminate form.
Using the formula $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x)-1]}$:
$\lim_{x \to 0} f(x) = e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{1+x}{1-x} - 1 \right)}$
$= e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{1+x - (1-x)}{1-x} \right)}$
$= e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{2x}{1-x} \right)}$
$= e^{\lim_{x \to 0} \frac{2}{1-x}}$
$= e^{\frac{2}{1-0}} = e^2$.
Therefore,$f(0) = e^2$.

(B) point in the domain of a function is said to have a non-removable discontinuity if the limit of the function as it approaches that point does not exist or is not equal to the function value,and it cannot be made continuous by redefining the function at that single point.
This is also known as an essential or jump discontinuity.
Therefore,the correct option is $B$.

(B) According to the Extreme Value Theorem,if a function $f$ is continuous on a closed and bounded interval $[a, b]$,then $f$ attains both a minimum value and a maximum value on that interval.
Therefore,the range of the function $f$ is the closed interval $[\text{Minimum } f, \text{Maximum } f]$.

(C) The function is defined as $f(x) = |x| + \frac{|x|}{x}$.
To check for continuity at $x = 0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
For $x < 0$,$|x| = -x$,so $f(x) = -x + \frac{-x}{x} = -x - 1$.
$\text{LHL} = \lim_{x \rightarrow 0^-} (-x - 1) = -1$.
For $x > 0$,$|x| = x$,so $f(x) = x + \frac{x}{x} = x + 1$.
$\text{RHL} = \lim_{x \rightarrow 0^+} (x + 1) = 1$.
Since $\text{LHL} \neq \text{RHL}$,the function is discontinuous at $x = 0$.
Note that $|x|$ is continuous at $x = 0$,but $\frac{|x|}{x}$ (the signum function) is discontinuous at $x = 0$.
Therefore,the function is discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there.
Thus,option $(c)$ is correct.

(A) For a function to be continuous at $x = -3$,we must have $\lim_{x \rightarrow -3} f(x) = f(-3)$.
Given $f(-3) = -\frac{5}{2}$.
Now,$\lim_{x \rightarrow -3} \frac{x^2+(a+3)x+(a+1)}{x+3} = -\frac{5}{2}$.
For the limit to exist,the numerator must be $0$ at $x = -3$:
$(-3)^2 + (a+3)(-3) + (a+1) = 0$
$9 - 3a - 9 + a + 1 = 0$
$-2a + 1 = 0 \implies a = \frac{1}{2}$.
Substituting $a = \frac{1}{2}$ into the numerator: $x^2 + 3.5x + 1.5 = (x+3)(x+0.5)$.
Thus,$\lim_{x \rightarrow -3} \frac{(x+3)(x+0.5)}{x+3} = \lim_{x \rightarrow -3} (x+0.5) = -3 + 0.5 = -2.5 = -\frac{5}{2}$.
This confirms $a = \frac{1}{2}$.
Now,we need to find $\lim_{x \rightarrow a} (x^2+x+1) = \lim_{x \rightarrow 1/2} (x^2+x+1) = (\frac{1}{2})^2 + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}$.

(D) function $f(x)$ is continuous at $x=a$ if $\lim _{x \rightarrow a^{+}} f(x) = \lim _{x \rightarrow a^{-}} f(x) = f(a)$.
Given $\lim _{x \rightarrow a^{+}} f(x)=p$,$\lim _{x \rightarrow a^{-}} f(x)=m$,and $f(a)=k$.
For continuity,we must have $p = m = k$.
Checking option $D$: If $p-m=0$,then $p=m$. If $p-k=0$,then $p=k$. Thus,$p=m=k$,which satisfies the condition for continuity at $x=a$.

(A) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x=0$ must be equal,i.e.,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = a$.
$1$. Calculate $LHL$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2(2x)}{x^2} = \lim_{x \to 0^-} 2 \left( \frac{\sin 2x}{2x} \right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
$2$. Calculate $RHL$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.
Rationalize the denominator:
$= \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 4+4 = 8$.
Since $LHL$ = $RHL$ = $8$,for the function to be continuous,$a$ must be $8$.

(B) For the function $f(x)$ to be continuous everywhere,it must be continuous at $x=0$ and $x=2$.
At $x=0$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1+\cos x) = 1+1 = 2$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (a-x) = a-0 = a$.
Since the function is continuous at $x=0$,$a = 2$.
At $x=2$:
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (a-x) = a-2 = 2-2 = 0$.
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2-b^2) = 4-b^2$.
Since the function is continuous at $x=2$,$0 = 4-b^2$,which implies $b^2 = 4$.
Therefore,$a^2+b^2 = 2^2 + 4 = 4+4 = 8$.

(C) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(0)$ must be equal.
$f(0) = a$.
$LHL$: $\lim_{x \to 0^-} (1+\sin x)^{\operatorname{cosec} x} = \lim_{x \to 0^-} (1+\sin x)^{1/\sin x} = e^1 = e$.
Since $f(x)$ is continuous at $x=0$,$a = e$.
$RHL$: $\lim_{x \to 0^+} \frac{e^{2/x}+e^{3/x}}{a e^{2/x}+b e^{3/x}}$.
Divide numerator and denominator by $e^{3/x}$:
$\lim_{x \to 0^+} \frac{e^{-1/x}+1}{a e^{-1/x}+b} = \frac{0+1}{a(0)+b} = \frac{1}{b}$.
Equating $RHL$ to $f(0)$: $\frac{1}{b} = a$.
Since $a = e$,we have $\frac{1}{b} = e$,which implies $b = \frac{1}{e} = e^{-1}$.
Therefore,$ab = e \times e^{-1} = 1$.

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) = 2$.
First,evaluate the right-hand limit $(RHL)$:
$\lim_{x \to 0^+} \frac{(e^{ax}-1) \log(1+x)}{\sin^2 x} = \lim_{x \to 0^+} \frac{(\frac{e^{ax}-1}{ax} \cdot ax) \cdot (\frac{\log(1+x)}{x} \cdot x)}{(\frac{\sin x}{x})^2 \cdot x^2} = \lim_{x \to 0^+} \frac{ax \cdot x}{x^2} = a$.
Since $f(0) = 2$,we have $a = 2$.
Next,evaluate the left-hand limit $(LHL)$:
$\lim_{x \to 0^-} \frac{\cos 4x - \cos bx}{\tan^2 x} = \lim_{x \to 0^-} \frac{-2 \sin(\frac{4x+bx}{2}) \sin(\frac{4x-bx}{2})}{\tan^2 x} = \lim_{x \to 0^-} \frac{-2 (\frac{4+b}{2}x) (\frac{4-b}{2}x)}{x^2} = -2 \cdot \frac{16-b^2}{4} = \frac{b^2-16}{2}$.
Since $f(0) = 2$,we have $\frac{b^2-16}{2} = 2$,which implies $b^2 - 16 = 4$,so $b^2 = 20$.
Finally,calculate $\sqrt{b^2 - a^2} = \sqrt{20 - 2^2} = \sqrt{20 - 4} = \sqrt{16} = 4$.

(C) For the function $f(x)$ to be continuous at $x = 1$,we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} e^{\frac{\sin a(x-[x])}{x-[x]}}$. Since $x < 1$,$[x] = 0$. Thus,$\lim_{x \to 1^-} e^{\frac{\sin ax}{x}} = e^{\sin a}$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{|x^2+x-2|}{x-1} = \lim_{x \to 1^+} \frac{|(x-1)(x+2)|}{x-1}$. Since $x > 1$,$(x-1) > 0$,so $|x-1| = x-1$. Thus,$\lim_{x \to 1^+} \frac{(x-1)(x+2)}{x-1} = \lim_{x \to 1^+} (x+2) = 3$.
$3$. Value at $x = 1$: $f(1) = b + 1$.
Equating $LHL$,$RHL$,and $f(1)$:
$e^{\sin a} = 3 \implies \sin a = \ln 3$.
$b + 1 = 3 \implies b = 2$.
Therefore,$b \sin a = 2 \ln 3 = \ln 3^2 = \ln 9$.

(D) For the function $f(x)$ to be continuous at $x = -1$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(-1)$ must be equal.
$1$. Calculate the value of the function at $x = -1$:
$f(-1) = \sin^{-1}[-1] = \sin^{-1}(-1) = -\pi / 2$.
$2$. Calculate the right-hand limit $(RHL)$ at $x = -1$:
$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \sin^{-1}[x]$.
Since $x$ approaches $-1$ from the right,$-1 < x < 0$,so $[x] = -1$.
Thus,$\lim_{x \to -1^+} \sin^{-1}(-1) = -\pi / 2$.
$3$. Calculate the left-hand limit $(LHL)$ at $x = -1$:
$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} k([x] + |x|)$.
Since $x$ approaches $-1$ from the left,$x < -1$,so $[x] = -2$ and $|x| = -x$.
Thus,$\lim_{x \to -1^-} k(-2 - x) = k(-2 - (-1)) = k(-2 + 1) = -k$.
$4$. Equate $LHL$ and $f(-1)$:
$-k = -\pi / 2 \implies k = \pi / 2$.

(D) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal,i.e.,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = a$.
$1$. Left-hand limit $(LHL)$:
$\lim_{x \to 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2(2x)}{x^2} = \lim_{x \to 0^-} 2 \left( \frac{\sin 2x}{2x} \right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
$2$. Right-hand limit $(RHL)$:
$\lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.
Rationalizing the denominator:
$\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16+0}+4 = 4+4 = 8$.
Since $LHL$ = $RHL$ = $8$,for the function to be continuous,$a$ must be equal to $8$.

(D) For $f(x)$ to be continuous at $x = -1$,we must have $\lim_{x \to -1} f(x) = f(-1) = k$.
The limit is $\lim_{x \to -1} \frac{(2-a)x^2 - (a+3b)x - 1}{x+1}$.
For the limit to exist,the numerator must be $0$ at $x = -1$.
Substituting $x = -1$: $(2-a)(-1)^2 - (a+3b)(-1) - 1 = 0 \implies 2 - a + a + 3b - 1 = 0 \implies 1 + 3b = 0 \implies b = -\frac{1}{3}$.
Substituting $b = -\frac{1}{3}$ into the numerator: $(2-a)x^2 - (a - 1)x - 1$.
This can be factored as $(x+1)((2-a)x - 1)$.
Thus,$\lim_{x \to -1} \frac{(x+1)((2-a)x - 1)}{x+1} = \lim_{x \to -1} ((2-a)x - 1) = (2-a)(-1) - 1 = -2 + a - 1 = a - 3$.
Since the function is continuous,$k = a - 3$.

(A) Given,$f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$
To check continuity at $x = 3$,we calculate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
$LHL$: $\lim_{x \to 3^{-}} f(x) = \lim_{x \to 3^{-}} \frac{2}{5-x} = \frac{2}{5-3} = 1$.
$RHL$: $\lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} (5-x) = 5-3 = 2$.
Also,$f(3) = 5-3 = 2$.
Since $\lim_{x \to 3^{-}} f(x) = 1$ and $\lim_{x \to 3^{+}} f(x) = 2$,the limit does not exist at $x = 3$.
Specifically,the left-hand limit is not equal to the value of the function at $x = 3$,therefore the function is left discontinuous at $x = 3$.

(A) The function is defined as $f(x) = \min \{x, x^2\}$.
By comparing $x$ and $x^2$,we find that $x^2 \leq x$ when $0 \leq x \leq 1$,and $x < x^2$ otherwise.
Thus,$f(x) = \begin{cases} x, & x < 0 \\ x^2, & 0 \leq x \leq 1 \\ x, & x > 1 \end{cases}$.
Checking continuity at $x = 0$: $\lim_{x \to 0^-} f(x) = 0$ and $\lim_{x \to 0^+} f(x) = 0^2 = 0$. Since $f(0) = 0$,it is continuous at $x = 0$.
Checking continuity at $x = 1$: $\lim_{x \to 1^-} f(x) = 1^2 = 1$ and $\lim_{x \to 1^+} f(x) = 1$. Since $f(1) = 1$,it is continuous at $x = 1$.
Now,checking differentiability: $f'(x) = \begin{cases} 1, & x < 0 \\ 2x, & 0 < x < 1 \\ 1, & x > 1 \end{cases}$.
At $x = 0$: Left derivative is $1$,right derivative is $2(0) = 0$. Since $1 \neq 0$,it is not differentiable at $x = 0$.
At $x = 1$: Left derivative is $2(1) = 2$,right derivative is $1$. Since $2 \neq 1$,it is not differentiable at $x = 1$.
Thus,$f(x)$ is continuous for all $x$.

(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{-}} f(x) = f(0) = \beta$.
First,evaluate the right-hand limit: $\lim_{x \to 0^{+}} \frac{\tan (\alpha + 1)x + \tan 2x}{x} = \lim_{x \to 0^{+}} \frac{\tan (\alpha + 1)x}{x} + \lim_{x \to 0^{+}} \frac{\tan 2x}{x} = (\alpha + 1) + 2 = \alpha + 3$.
Thus,$\beta = \alpha + 3$.
Next,evaluate the left-hand limit: $\lim_{x \to 0^{-}} \frac{\sin 3x - \tan 3x}{x^{3}}$. Using the Taylor series expansion $\sin 3x = 3x - \frac{(3x)^3}{6} + O(x^5)$ and $\tan 3x = 3x + \frac{(3x)^3}{3} + O(x^5)$,we get:
$\lim_{x \to 0^{-}} \frac{(3x - \frac{27x^3}{6}) - (3x + \frac{27x^3}{3})}{x^3} = \lim_{x \to 0^{-}} \frac{-\frac{9}{2}x^3 - 9x^3}{x^3} = -\frac{9}{2} - 9 = -\frac{27}{2}$.
So,$\beta = -\frac{27}{2}$.
Substituting $\beta$ into the first equation: $-\frac{27}{2} = \alpha + 3 \implies \alpha = -\frac{27}{2} - 3 = -\frac{33}{2}$.
Finally,$|\alpha| + |\beta| = |-\frac{33}{2}| + |-\frac{27}{2}| = \frac{33}{2} + \frac{27}{2} = \frac{60}{2} = 30$.

(B) For the function $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0^-} f(x) = f(0) = \lim_{x \rightarrow 0^+} f(x)$.
First,calculate the left-hand limit:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (1 + |\sin x|)^{a/|\sin x|}$.
Since this is of the form $1^\infty$,we use the formula $\lim_{x \rightarrow c} (1 + g(x))^{h(x)} = e^{\lim_{x \rightarrow c} g(x)h(x)}$.
$\lim_{x \rightarrow 0^-} f(x) = e^{\lim_{x \rightarrow 0} |\sin x| \cdot \frac{a}{|\sin x|}} = e^a$.
Next,calculate the right-hand limit:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} e^{\frac{\tan 2x}{\tan 3x}} = e^{\lim_{x \rightarrow 0} \frac{\tan 2x}{2x} \cdot \frac{3x}{\tan 3x} \cdot \frac{2x}{3x}} = e^{1 \cdot 1 \cdot 2/3} = e^{2/3}$.
Since $f(0) = b$,we equate the limits:
$e^a = b = e^{2/3}$.
Therefore,$a = 2/3$ and $b = e^{2/3}$.

(C) For a function $f(x)$ to be continuous at $x=0$,the condition $\lim_{x \rightarrow 0} f(x) = f(0)$ must hold.
We calculate the limit as follows:
$f(0) = \lim_{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$
To evaluate this limit,we rationalize the numerator:
$f(0) = \lim_{x \rightarrow 0} \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}$
$f(0) = \lim_{x \rightarrow 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)}$
$f(0) = \lim_{x \rightarrow 0} \frac{x}{x(\sqrt{1+x}+1)}$
$f(0) = \lim_{x \rightarrow 0} \frac{1}{\sqrt{1+x}+1}$
Substituting $x=0$:
$f(0) = \frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2}$